Math homework help (8th grade)

[episodic]

The graphic didn't work.
It looks like stairs made with boxes.

1st has 1
2nd has 3
3rd has 6th

Homework was to compute the 50th iteration. . .


I'm helping a kid here. I know the function is x(x+1) divided by 2

I'm trying to explain how he gets it though. I just happened to remember triangular numbers and the pattern myself. . .I'm rusty on this stuff.

How would you explain it?

x y
1 1
2 3
3 6
4 10
5 15

I understand it is +2, +3, +4, etc. . .

I'm trying to get em to deeply understand it. . .


remember the grade level here - no college level explanations.

 

[imported_Tick]

If you graph it, it's clearly a parabola. Then just compute the equation of a parabola. You'll wind up with your equation, and ta-da.

 

[Q]

Tick's right, have him graph it then find the equation using the graph. (Rise over run)

 

[Syppion]

Are you sure he's supposed to get it in that form?

At first glance I would have said

Xn = n+(n-1)+(n-2)+...+1

or

Xn = SUM(j) | j = 1->n
(sum of j evaluated from j = 1 to j = n) - forum syntax is a bit ugly.

Anyhow, the context will have to tell which form it should be in.

 

[Suspicious-Teach8788]

Summation... 1 + 2 + 3 + 4 + 5 + 6...

Sigma X where x goes from 1 to 50. Remember your summation formulas! What is it? 1/2 * n * (n-1) ? Algebra 2 was too long ago.

 

[episodic]

it is supposed to be in y = form

Not quite sure how you compute the equation from the slope of a parabola???? I can compute slopes of linear equations. . .

dlerium yea it is

n(n+1)
--------
2

I'm just trying to get him to understand it more deeply. He knows why it works. . .I can't figure an easy way for em to come up with it on their own just looking at the table. . .

 

[videogames101]

Originally posted by: episodic
it is supposed to be in y = form

Not quite sure how you compute the equation from the slope of a parabola???? I can compute slopes of linear equations. . .

isn't it just y=x(x+1) and just put the x(x+1) over 2?

 

[Q]

Originally posted by: episodic
Not quite sure how you compute the equation from the slope of a parabola???? I can compute slopes of linear equations. . .

Oh wait, you're right.

That is a hard problem, why the heck is he getting that in the 8th grade?

 

[episodic]

ok, i'm dumb. I had the float set to 0 on a ti-84. It just dawned on me in my sleep deprived state that a parabola can't have an r squared value of 1 with a linear equation. . . sheesh. . . The quadratic r squared value was 1 too. It wasn't working right because I had the calculator set to 0 float. . . (rounding off to the nearest whole number ) we got it now, and the understanding is getting better.

 

[videogames101]

Originally posted by: Quintox

Originally posted by: episodic
Not quite sure how you compute the equation from the slope of a parabola???? I can compute slopes of linear equations. . .

Oh wait, you're right.

That is a hard problem, why the heck is he getting that in the 8th grade?

uh, the anwser is simple, the question here is how to deeply explain, it, which would require a collage level understanding of mathmatics. My advice, just forget about the possibility of understanding it, but i'll try to remember what i learned abvout this...

 

[episodic]

Originally posted by: Quintox

Originally posted by: episodic
Not quite sure how you compute the equation from the slope of a parabola???? I can compute slopes of linear equations. . .

Oh wait, you're right.

That is a hard problem, why the heck is he getting that in the 8th grade?

Pre-AP Algebra I Part B for high school credit. . .

 

[Tiamat]

y = 0.5x^2 + 0.5x

Plug in x = 50, calculate y

This is with excel and is the wrong way for an 8th grader

The right way is to notice that the 50th iteration is 50 more than the 49th iteration. The 49th iteration is 25 times 49.

 

[episodic]

Originally posted by: Tiamat
y = 0.5x^2 + 0.5x

Plug in x = 50, calculate y

yea, got that immediately after I figured the goof with my graphing calculator having the float set to 0 haha

 

[Q]

Believe me, classes for "high school credit" aren't worth it. It's not very important IMO. I didn't have any classes for HS credit and I made it out fine and am in a very good liberal arts school this year (also got accepted to Clemson, USC, etc)

 

[episodic]

Tell me if this works (for deep understanding)

xx
*x
**

The box above is the 2nd iteration. The asteriks are the problem (2nd iteration, 3 blocks). The x's are the mirror image of the problem forming a rectangle. Finding area is found by length times width.

the bottom is 'x' long
The side will be x+1 long every iteration if you make a mirror image

for instance
xxx
*xx
**x
***

That is the next iteration
The bottom is x long and again the side is x+1 one long

Area says that x times (x+1) will give you the area

so above is the 3rd iteration so 3(3+1) = 12

It follows that the blocks are just half of 12 (how do you go from 12 to the 6 in the iteration) - divide by 2

So it follows all numbers of blocks in each iteration is really just the area of a rectangle x long and x+1 high divided into a right triangle (divided by 2). . .

Does that work?

 

[videogames101]

This porblem is more about learning how triangular numbers work, so that area thing seems kinda off...

 

[Tiamat]

See above edit. This question is a standard GRE question and this is how you solve it on the GRE.



The right way is to notice that the 50th iteration is 50 more than the 49th iteration. The 49th iteration is 25 times 49.

(This is because you can see that the 4th iteration is 4 more than the 3rd iteration. The 3rd iteration is twice the iteration number because it is the 2nd odd number in the sequence.

49 is the 25th odd number of the sequence...

Therefore, the only calculation in this problem is 49*25 + 50.

Episodic - the best way for you to tutor the kid is to read a GRE math help. In college, we have learned to use MS Excel and graphic calculators which we have grown too dependent on.

 

[episodic]

Any comment on what I did above with the area model?

 

[Tiamat]

Originally posted by: episodic
Any comment on what I did above with the area model?

I take that back.. I have to study what you said a bit more. My method is pretty sound though.

I cant say that I would solve the problem as you proposed. My brain doesnt work like that The problem when I see it at quick glance is a problem where you find the pattern and apply the pattern. Thus, what I proposed.

Kind of like the problem: Add all numbers inclusive in these ranges 9-28, 12-29. Which range has the highest sum? Solve this problem in 10 seconds.

The trick to this problem is overlap. The problem is actually 9+10+11 vs 29

 

[QED]

Am I missing something here? These are just triangular numbers, and the easiest way to discover their associated formula is to add the numbers up twice. You can then pair the addends together in such a way that each pair sum to x+1 (by pairing 1 and n, 2 and x-1, 3 and x-2,etc). You will have x pairs, so this double sum is simply equal to x * (x+1).



For instance, for x = 5, you have y = 1 + 2 + 3 + 4 +5 . Hence,
y + y = (1 + 2 + 3 + 4 + 5 ) + (1 + 2 + 3 + 4 + 5)
= (1 + 2 + 3 + 4 + 5 ) + (5 + 4 + 3 + 2 + 1)
= (1+5) + (2+4) + (3 + 3) + (4 + 2) + (5 + 1)
= 6 + 6 + 6 + 6 + 6
= 6 * 5, hence
2y = 6 * 5, or y = 5 * 6 / 2.

Or, you can visualize this with the stairs made of boxes. Make a copy of the stairs, flip them around 180 degrees, and fit them against the original stairs. You will now have a rectangle of boxes with x rows and x+1 columns. Hence, the total number of boxes for BOTH stairs is x * (x+1), so the number of boxes for just one of the set of stairs is x * (x+1) / 2.

 

[Madwand1]

Originally posted by: episodic
Any comment on what I did above with the area model?

I think it's fine, but I personally find the part about the "mirror image" to be a bit unclear/unintuitive -- how exactly is the "mirror image" defined and growing, and how do you know it'll work for the nth iteration? To clear this up, the "mirror image" can be defined as a set of "blank" blocks which are growing just as the problem blocks do, but vertically instead of horizontally. The nth step adds one more level up of n more blank blocks at the top. This is well-defined, maintains symmetry, and shows the same growth rate as the counting blocks.

There are other solutions, including a famous one attributed to Gauss in a childhood story. The problem amounts to adding the sequence 1+2+3... +48+49+50. Instead of adding left to right, consider the pairs

1 + 50 = 51
2 + 49 = 51
3 + 48 = 51
etc.

The number on the left iterator is growing by one, and the number on the right iterator is shrinking by one, so the sum is constant.

For an even numbered sequence, there are exactly n/2 such sums. So the formula n/2 (n+1) is directly derived. This actually solves the given problem.

Here's a graphical "blocks" version of this solution. http://www.freewebs.com/keiich...uki/child/suuretu.html

For an odd numbered sequence, there are (n-1)/2 such sums plus the middle term. The middle term is (n+1)/2, so the total is (n-1)/2 * (n+1) + (n+1)/2 = ((n-1)/2 + 1/2) * (n+1) = n/2 (n+1), thus this holds in all cases.

Another solution based on "blocks" would build squares instead of rectangles. Here you'd have n^2 total blocks, and sum (n-1) empty blocks (you need to draw the blocks, etc. to see these), so a formula would be sum (n) = n^2 - sum (n-1)

Given sum (n) = sum (n-1) + n by definition, you could derive sum (n) = n/2(n+1) by putting together the two formula and solving for sum (n-1) and then substituting the next term, n, for the solution for sum (n).

But this gets unintuitive. I think Gauss' solution is probably the best, being easiest to remember and generalize.

Edit: I didn't see QED's post when I wrote the above. His doubling of the entire sequence is even clearer than the "even" and "odd" solutions given here.